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Many physical processes are exponential in nature. A typical way of cleaning a tank of contaminated water is to run in clean water at a constant rate, stir, and let the mixture run out at the same rate. Suppose there are initially 100 pounds of a contaminant in a large tank of water. Assume that the cleaning method described above removes 20% of the remaining contaminant each hour. How much contaminant is removed during the first three hours

Respuesta :

Answer:

The amount of the contaminant removed during the first three hours will be 48.8 pounds.

Step-by-step explanation:

There, in a large tank of water, are initially 100 pounds of a contaminant.

Also given that the cleaning method removes 20% of the remaining contaminant each hour.

So, after the first hour, the amount of the contaminant will remain by [tex]100(1 - \frac{20}{100}) = 80[/tex] pounds.

Now, after the second hour, the amount of the contaminant will remain by

[tex]80(1 - \frac{20}{100}) = 64[/tex] pounds.

Finally, after the third hour, the amount of the contaminant will remain by [tex]64(1 - \frac{20}{100}) = 51.2[/tex] pounds.

Therefore, the amount of the contaminant removed during the first three hours will be (100 - 51.2) = 48.8 pounds. (Answer)

The amount of the contaminant removed during the first three hours will be 48.8 pounds.

Calculation of the amount of the contaminant:

Since there are initially 100 pounds of a contaminant in a large tank of water. Assume that the cleaning method described above removes 20% of the remaining contaminant each hour.

So,  

= 100(1 - 0.20)

= 80

And,

= 80(1 - 0.20)

= 64

And,

= 0.64(1 - 0.20)

= 51.2

Now the amount should be

= 100 - 51.2

= 48.8

Hence, The amount of the contaminant removed during the first three hours will be 48.8 pounds.

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